Does block product(=inner product) of the normal subgroups are commutative?

Question

Let the $N, M \lhd G$ $s.t. M \cap N = \{ e\}$

• $e$ means identitiy of the group and N, M are normal subgroups of the G.

Say $\forall n \in N$ and $\forall m \in M$, Then I've already knew that $nm=mn$.

My question is

Does the group $NM (\lhd G)$ is commutative group or not ?

Intutitvely, It looks like a true since the $NM=MN$

But, When I tried the proof the following problems happened.

$pf)$ $\forall n_1, n_2 \in N$ and $\forall m_1, m_2 \in M$

Want to show $(n_1 m_1)(n_2m_2)$ = $(n_2 m_2)(n_1m_1)$

$(L.H.S.)$ = $(n_1n_2)(m_1m_2)$ since $nm=mn$

$(R.H.S)$ = $(n_2 n_2)(m_2m_1)$ since $nm=mn$

One thing sure that We can't say $(n_1n_2) =(n_2n_1)$.

Therefore

If the statement is true, What next process I need for the $(n_1n_2)(m_1m_2) =(n_2n_1)(m_2m_1)$ ?

If the statement is false Might be find the counterexaples the Normal subgroups but not commutative. But I couldn't find those are.

Answer 1

Consider if $G=M\times N$ with $M$ and $N$ non-abelian. Then any element $(m,e_N)$ commutes with any element $(e_M,n)$, but $G$ is non-abelian.